Simplify; express your answer in exponential form. Assume $y\neq 0, t\neq 0$. $\dfrac{{(y^{-1})^{5}}}{{(y^{2}t^{3})^{-2}}}$
Explanation: To start, try working on the numerator and the denominator independently. In the numerator, we have ${y^{-1}}$ to the exponent ${5}$ . Now ${-1 \times 5 = -5}$ , so ${(y^{-1})^{5} = y^{-5}}$ In the denominator, we can use the distributive property of exponents. ${(y^{2}t^{3})^{-2} = (y^{2})^{-2}(t^{3})^{-2}}$ Simplify using the same method from the numerator and put the entire equation together. $\dfrac{{(y^{-1})^{5}}}{{(y^{2}t^{3})^{-2}}} = \dfrac{{y^{-5}}}{{y^{-4}t^{-6}}}$ Break up the equation by variable and simplify. $\dfrac{{y^{-5}}}{{y^{-4}t^{-6}}} = \dfrac{{y^{-5}}}{{y^{-4}}} \cdot \dfrac{{1}}{{t^{-6}}} = y^{{-5} - {(-4)}} \cdot t^{- {(-6)}} = y^{-1}t^{6}$.